\begin{answer}
    Note
    $$
    \begin{aligned}
        E_{y\sim p(y;\theta)}[\nabla_{\theta'}\log p(y;\theta')|_{\theta' = \theta}] &= \int_{-\infty}^{\infty}(\nabla_{\theta'}\log p(y;\theta')|_{\theta' = \theta}) p(y;\theta)dy \\
        &= \int_{-\infty}^{\infty}p(y;\theta)\frac{1}{p(y;\theta)} \nabla_\theta p(y;\theta)dy\\
        &= \nabla_\theta \int_{-\infty}^{\infty}p(y;\theta)dy = 0
    \end{aligned}
    $$

\end{answer}
